Chemistry 9th Edition

Published by Cengage Learning
ISBN 10: 1133611095
ISBN 13: 978-1-13361-109-7

Chapter 5 - Gases - Exercises - Page 237: 71


$33.2\text{ grams $KrCl_4$}$

Work Step by Step

Reaction: $$Kr+2Cl_2\Rightarrow KrCl_4$$ Find the limiting reactant: $PV=nRT$ $0.5*15.0=n(0.08206)(623)$ $n=0.147\text{ mol Kr}$ $1.5*15.0=n*0.08206*623$ $n=0.4401\text{ mol $Cl_2$}$ Kr is the limiting reactant. $0.147\text{ mol Kr}*\frac{1\text{ mol $KrCl_4$}}{1\text{ mol Kr}}*\frac{225.6100\text{ grams $KrCl_4$}}{1\text{ mol $KrCl_4$}}=33.2\text{ grams $KrCl_4$}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.